Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n - 6}{n + 9} \times \dfrac{-6n - 42}{n^2 + n - 42} $
Answer: First factor the quadratic. $a = \dfrac{n - 6}{n + 9} \times \dfrac{-6n - 42}{(n - 6)(n + 7)} $ Then factor out any other terms. $a = \dfrac{n - 6}{n + 9} \times \dfrac{-6(n + 7)}{(n - 6)(n + 7)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (n - 6) \times -6(n + 7) } { (n + 9) \times (n - 6)(n + 7) } $ $a = \dfrac{ -6(n - 6)(n + 7)}{ (n + 9)(n - 6)(n + 7)} $ Notice that $(n + 7)$ and $(n - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -6\cancel{(n - 6)}(n + 7)}{ (n + 9)\cancel{(n - 6)}(n + 7)} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $a = \dfrac{ -6\cancel{(n - 6)}\cancel{(n + 7)}}{ (n + 9)\cancel{(n - 6)}\cancel{(n + 7)}} $ We are dividing by $n + 7$ , so $n + 7 \neq 0$ Therefore, $n \neq -7$ $a = \dfrac{-6}{n + 9} ; \space n \neq 6 ; \space n \neq -7 $